Question: Graph this system of equations and solve. $10x+5y = -25$ $6x+10y = 20$ 1 2 3 4 5 6 7 8 9 10 \llap{-}2 \llap{-}3 \llap{-}4 \llap{-}5 \llap{-}6 \llap{-}7 \llap{-}8 \llap{-}9 \llap{-}10 1 2 3 4 5 6 7 8 9 10 \llap{-}2 \llap{-}3 \llap{-}4 \llap{-}5 \llap{-}6 \llap{-}7 \llap{-}8 \llap{-}9 \llap{-}10 Click and drag the points to move the lines.
Convert the first equation, $10x+5y = -25$ , to slope-intercept form. $y = -2 x - 5$ The y-intercept for the first equation is $-5$ , so the first line must pass through the point $(0, -5)$ The slope for the first equation is $-2$ . Remember that the slope tells you rise over run. So in this case for every $2$ positions you move down (because it's negative) $1$ position to the right. $2$ positions down from $(0, -5)$ is $(1, -7)$ Graph the blue line so it passes through $(0, -5)$ and $(1, -7)$ Convert the second equation, $6x+10y = 20$ , to slope-intercept form. $y = -\dfrac{3}{5} x + 2$ The y-intercept for the second equation is $2$ , so the second line must pass through the point $(0, 2)$ The slope for the second equation is $-\dfrac{3}{5}$ . Remember that the slope tells you rise over run. So in this case for every $3$ positions you move down (because it's negative) $5$ positions to the right. $3$ positions down from $(0, 2)$ is $(5, -1)$ Graph the green line so it passes through $(0, 2)$ and $(5, -1)$ The solution is the point where the two lines intersect. The lines intersect at $(-5, 5)$.